Past exam of the statistics course of the University of Oxford / 2023 / 2

For a critical region $C$, the size of a statistical test of the simple null hypothesis $P_0$ is $\alpha =P_0(\mathbf{X}\in C)$, the Type I error probability. Its statistical power against the simple alternative hypothesis $P_1$ is $P_1(\mathbf{X}\in C)$, the probability of correctly rejecting that null. For a randomized test function $\phi (\mathbf{x})\in [0,1]$, replace these by ${\mathbb{E}}_0\phi$ and ${\mathbb{E}}_1\phi$.

The Neyman-Pearson lemma states that a test which rejects where the joint likelihood ratio $L_1/L_0$ is largest is a most powerful test of its size: its power is at least that of any test with no larger size. More precisely, choose a threshold $k$ and reject if $L_1/L_0>k$, accept if it is smaller, and if necessary randomize on equality to attain the specified size.

For an independent and identically distributed statistical sample, takeThe $Y_i$ are independent and identically distributed random variables under either simple hypothesis, and the log joint ratio is their sum. Thus the nonrandomized threshold form can be writtenwith $c=-\log k$, using either strict or non-strict inequality when the boundary has probability zero. The usual finite-log expression assumes positive common support; zero densities can instead be handled by the extended likelihood ratio on the relevant supports. Continuous observations alone do not ensure a continuous likelihood-ratio statistic: ties may require the boundary randomization in the lemma. This qualification preserves the exact-size formulation.

The joint likelihood ratio in favor of the alternative isIt is increasing in the product, so the Neyman-Pearson lemma rejects for large product, equivalently small $S=-{\sum }_i\log x_i$. Under $H_0$, the transformation $Z_i=-\log X_i$ givesTherefore the $Z_i$ are independent exponential random variables of rate two, $S\sim \mathrm{Gamma}(n,\text{rate }2)$, and $4S\sim {\chi }_{2n}^2$. An exact size-$\alpha$ critical region isEquivalently, reject when the geometric mean satisfies $g\ge e^{-c_{\alpha }/n}$. The boundary distribution is continuous, so no randomization is needed here.

Under $H_0$, each $Z_i=-\log X_i$ has expected value $1/2$ and variance $1/4$. The central limit theorem therefore approximatesFor the supplied data this standardized value isUsing the provided approximation $\Phi (1.6)\approx 0.95$, the lower-tail 5% rule rejects when $S\le 12.5-1.6(2.5)=8.5$. The observed $6.25$ is in this region, with approximate p-value $\Phi (-2.5)\approx 0.005$. Reject $H_0:\theta =2$ at the 5% level in favor of the specified alternative $\theta =3$. The unusually large geometric mean is consistent with the direction of that alternative. This is evidence in this comparison, not a proof that the parameter is exactly three.